Variance Using Mean of Squares

For $$N = 10$$ observations:

$$\bar{x}_{10} = 5 \implies \sum_{i=1}^{10} x_i = 10 \times 5 = 50$$

$$\sigma_{10}^2 = 7 \implies \frac{1}{10}\sum_{i=1}^{10} x_i^2 - (\bar{x}_{10})^2 = 7 \implies \frac{1}{10}\sum_{i=1}^{10} x_i^2 - 25 = 7 \implies \sum_{i=1}^{10} x_i^2 = 320$$

For the first 8 observations ($$N = 8$$):

$$\bar{x}_{8} = 4 \implies \sum_{i=1}^{8} x_i = 8 \times 4 = 32$$

$$\sigma_{8}^2 = 3.5 \implies \frac{1}{8}\sum_{i=1}^{8} x_i^2 - (\bar{x}_{8})^2 = 3.5 \implies \frac{1}{8}\sum_{i=1}^{8} x_i^2 - 16 = 3.5 \implies \sum_{i=1}^{8} x_i^2 = 156$$

Calculating relations for $$x_9$$ and $$x_{10}$$:

$$\sum_{i=1}^{10} x_i - \sum_{i=1}^{8} x_i = x_9 + x_{10} \implies 50 - 32 = x_9 + x_{10} \implies x_9 + x_{10} = 18 \quad \text{--- (1)}$$

$$\sum_{i=1}^{10} x_i^2 - \sum_{i=1}^{8} x_i^2 = x_9^2 + x_{10}^2 \implies 320 - 156 = x_9^2 + x_{10}^2 \implies x_9^2 + x_{10}^2 = 164 \quad \text{--- (2)}$$

$$18^2 = 164 + 2x_9x_{10} \implies 324 = 164 + 2x_9x_{10} \implies x_9x_{10} = 80 \quad \text{--- (3)}$$

Solving the quadratic equation $$t^2 - (x_9+x_{10})t + x_9x_{10} = 0$$:

$$t^2 - 18t + 80 = 0 \implies (t-8)(t-10) = 0 \implies t = 8, 10$$

Since $$x_9 < x_{10}$$: $$x_9 = 8, \quad x_{10} = 10$$

$$3x_9 + 2x_{10} = 3(8) + 2(10) = 24 + 20 = 44$$

Given:

Number of observations,

$$n=10$$

Mean,

$$\bar x=15$$

Variance,

$$\sigma^2=16$$

Using the formula

$$\sigma^2=\frac{\sum x_i^2}{n}-\bar x^2,$$

we get

$$16=\frac{\sum x_i^2}{10}-15^2$$

$$16=\frac{\sum x_i^2}{10}-225$$

$$\frac{\sum x_i^2}{10}=241$$

$$\sum x_i^2=2410$$

Also,

$$\sum x_i=10\times15=150$$

Now after adding the new observation $$k,$$ the mean becomes $$16.$$

Hence,

$$\frac{150+k}{11}=16$$

$$150+k=176$$

$$k=26$$

Now the new sum of squares becomes

$$2410+26^2$$

$$=2410+676$$

$$=3086$$

Variance of the $$11$$ observations:

$$\sigma^2=\frac{3086}{11}-16^2$$

$$=\frac{3086}{11}-256$$

$$=\frac{3086-2816}{11}$$

$$=\frac{270}{11}$$

$$\approx24.54$$

Closest integeris

$$\boxed{25}$$

Given mean $$=4$$ for $$10$$ observations,

$$\sum x_i=40$$

Given variance

$$=9,$$

$$9=\frac{\sum x_i^2}{10}-16$$

$$\sum x_i^2=250$$

After adding $$4,$$

new sum becomes $$44,$$

hence new mean remains $$4$$

Now,

$$x_1^2+x_2^2=50$$ is replaced by $$3^2+5^2=34$$

Also include

$$4^2=16$$

Hence new sum of squares:

$$250-50+34+16=250$$

Therefore,

$$\sigma^2=\frac{250}{11}-16=\frac{74}{11}$$

Thus,

$$11\sigma^2=74$$